Question

The integer n for which $\lim_{x \rightarrow 0}\frac{\left( \cos x - 1 \right)\left( \cos x - e^{x} \right)}{x^{n}}$ is a finite non-zero number is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

Give that, $\lim_{x \rightarrow 0}\frac{\left( \cos x - 1 \right)\left( \cos x - e^{x} \right)}{x^{n}}$

= finite non zero number

= $\lim_{x \rightarrow 0}\frac{\left( \cos x - 1 \right)\left( 1 + \cos x \right)\left( e^{x} - \cos x \right)}{x^{n}\left( 1 + \cos x \right)}$

= $\lim_{x \rightarrow 0}\left( \frac{\sin^{2}x}{x^{2}} \right).\left( \frac{\left( e^{x} - \cos x \right)}{x^{n - 2}} \right).\left( \frac{1}{1 + \cos x} \right)$

= $1^{2}.\frac{1}{2}\lim_{x \rightarrow 0}\frac{\left\lbrack 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + .....\infty \right\rbrack - \left\lbrack 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + .....\infty \right\rbrack}{x^{n - 2}}$

=$\frac{1}{2}\lim_{x \rightarrow 0}\frac{x\left( 1 + x + \frac{x^{2}}{3!} + \frac{2x^{3}}{4!} + .....\infty \right)}{x^{n - 2}}$

= $\frac{1}{2}\lim_{x \rightarrow 0}\frac{x\left( 1 + x + \frac{x^{2}}{3!} + \frac{2x^{3}}{4!} + .....\infty \right)}{x^{n - 3}}$

For this limit to be finite n − 3 = 0

⇒ n = 3